POJ1753(广搜)

广搜,位运算存储状态记录是否访问过。

#include 
#include 
#include 

using namespace std;

struct S
{
	int A[16];
	int pos;
};

int get_num(S s)
{
	int i;
	int rec=0;
	for (i=0;i<=15;i++)
		if (s.A[i]==0)
			rec=rec<<1;
		else
			rec=(rec<<1)+1;
	return rec;
}

int main()
{
	S s;
	int i,j;
	char str[5];
	int record[66000]={0};
	int counter;
	int count=0;
	queue Q;

	for (i=0;i<=3;i++)
	{
		scanf("%s",str);
		for (j=0;j<=3;j++)
		{
		  if (str[j]=='b')
			  s.A[count++]=0;
		  else
			  s.A[count++]=1;
		}
	}

	int flag=0;
	int num;
	S temp;
	s.pos=0;
  Q.push(s);
  //printf("%d\n",get_num(s));
  record[get_num(s)]=1;
  while (!Q.empty())
  {
  	num=get_num(Q.front());
  	if (num==0 || num==65535)
  	{
  		counter=Q.front().pos;
  		flag=1;
  		break;
  	}

  	for (i=0;i<=15;i++)
  	{
    	temp=Q.front();
  		if (i-4>=0)
  			temp.A[i-4]=!temp.A[i-4];
  		if (i+4<=15)
  			temp.A[i+4]=!temp.A[i+4];
  		if ( (i+1)%4!=0 )
  			temp.A[i+1]=!temp.A[i+1];
  		if ( i%4!=0 )
  			temp.A[i-1]=!temp.A[i-1];
  		temp.A[i]=!temp.A[i];

  		num=get_num(temp);
  		if (!record[num])
  		{
  			temp.pos=Q.front().pos+1;
  	  	Q.push(temp);
  		  record[num]=1;
  		}
  	}
  	Q.pop();
  }

  if (flag)
  	printf("%d\n",counter);
  else
  	printf("Impossible\n");
	return 0;
}

转载的高斯消元法,更高效的解法:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
//对2取模的01方程组
const int MAXN = 300;
//有equ个方程,var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元(多解枚举自由变元可以使用)
int free_num;//自由变元的个数

//返回值为-1表示无解,为0是唯一解,否则返回自由变元个数
int Gauss()
{
    int max_r,col,k;
    free_num = 0;
    for(k = 0, col = 0 ; k < equ && col < var ; k++, col++)
    {
        max_r = k;
        for(int i = k+1;i < equ;i++)
        {
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        }
        if(a[max_r][col] == 0)
        {
            k--;
            free_x[free_num++] = col;//这个是自由变元
            continue;
        }
        if(max_r != k)
        {
            for(int j = col; j < var+1; j++)
                swap(a[k][j],a[max_r][j]);
        }
        for(int i = k+1;i < equ;i++)
        {
            if(a[i][col] != 0)
            {
                for(int j = col;j < var+1;j++)
                    a[i][j] ^= a[k][j];
            }
        }
    }
    for(int i = k;i < equ;i++)
        if(a[i][col] != 0)
            return -1;//无解
    if(k < var) return var-k;//自由变元个数
    //唯一解,回代
    for(int i = var-1; i >= 0;i--)
    {
        x[i] = a[i][var];
        for(int j = i+1;j < var;j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}
int n;
void init()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    equ = n*n;
    var = n*n;
    for(int i = 0;i < n;i++)
        for(int j = 0;j < n;j++)
        {
            int t = i*n+j;
            a[t][t] = 1;
            if(i > 0)a[(i-1)*n+j][t] = 1;
            if(i < n-1)a[(i+1)*n+j][t] = 1;
            if(j > 0)a[i*n+j-1][t] = 1;
            if(j < n-1)a[i*n+j+1][t] = 1;
        }
}
const int INF = 0x3f3f3f3f;
int solve()
{
    int t = Gauss();
    if(t == -1)
    {
        return INF;
    }
    else if(t == 0)
    {
        int ans = 0;
        for(int i = 0;i < n*n;i++)
            ans += x[i];
        return ans;
    }
    else
    {
        //枚举自由变元
        int ans = INF;
        int tot = (1<= 0;j--)
            {
                int idx;
                for(idx = j;idx < var;idx++)
                    if(a[j][idx])
                        break;
                x[idx] = a[j][var];
                for(int l = idx+1;l < var;l++)
                    if(a[j][l])
                        x[idx] ^= x[l];
                cnt += x[idx];
            }
            ans = min(ans,cnt);
        }
        return ans;
    }
}
char str[10][10];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    n = 4;
    for(int i = 0;i < 4;i++)
        scanf("%s",str[i]);
    init();
    for(int i = 0;i < 4;i++)
        for(int j = 0;j < 4;j++)
        {
            if(str[i][j] == 'b')a[i*4+j][16] = 0;
            else a[i*4+j][16] = 1;
        }
    int ans1 = solve();
    init();
    for(int i = 0;i < 4;i++)
        for(int j = 0;j < 4;j++)
        {
            if(str[i][j] == 'b')a[i*4+j][16] = 1;
            else a[i*4+j][16] = 0;
        }
    int ans2 = solve();
    if(ans1 == INF && ans2 == INF)
        printf("Impossible\n");
    else printf("%d\n",min(ans1,ans2));
    return 0;
}


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